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EXPAND sqrt{/frac{1-x}{1+x}}

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\[\Huge \sqrt{\frac{1-x}{1+x}}\]
up to power x^2
i get \[1-x+\frac{x^2}{2} - \frac{x^3}{8} - \frac{3x^4}{64}\]

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Other answers:

wolfram got \[-\frac{3 x^5}{8}+\frac{3 x^4}{8}-\frac{x^3}{2}+\frac{x^2}{2}-x+1\]
Unfortunately i forgot the formula! D:
Is it a maclaurin series?
you can try that
I want to know what is meant by Expand, since that's the only thing that comes to mind.
you can use this \[(x+a)^n=\sum ^{n}_{ k=0} \left(\begin{matrix}n \\ k\end{matrix}\right) x^k a^{(n-k)}\]
or \[(1+x)^n = 1+nx+\frac{(n)(n-1)}{2!}(x^2)+...\]
That works for \(n\in \mathbb{N}\).
In this case you have a square root...
still can, watch
\[\sqrt{\frac{1-x}{1+x}} \\ \\ \frac{\sqrt{1-x}}{\sqrt{1+x}} \\ \\ (1-x)^\frac{1}{2}(1+x)^{-\frac{1}{2}}\]
expand separately
mix them by multiplying together
Yeah, my problem is with \(1/2 \notin \mathbb{N}\)
I don't think it should be natural number
i mean real number will do
How can you have summations or binomial coefficients with rational numbers?
interval of x is ?
the formula list says can use
1 Attachment
we use the second one (1+x)^n look at the History section
or maybe its \[-\frac{3 x^5}{8}+\frac{3 x^4}{8}-\frac{x^3}{2}+\frac{x^2}{2}-x+1\]
based from the formula as long as its in the from ("1"+x)^n should be fine to expand, i tried it
from \[(1-x)^{1/2} \\ \\ n=-1/2, x=-x(for formula) \\ \\ [1-\frac{x}{2}+\frac{(\frac{1}{2})(\frac{1}{2}-1)}{2}(x^2)] \\ \\ 1-\frac{x}{2}-\frac{x^2}{8}\]

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